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Electrochemistry Common Student Misconceptions • Students often have trouble balancing redox equations. Lecture Outline 1. Oxidation States • Chemical reactions in which the oxidation state of one or more substances change are called oxidation-reduction reactions (redox reactions). • Recall: • Oxidation involves loss of electrons (OIL). • Reduction involves gain of electrons (RIG). • Electrochemistry is the branch of chemistry that deals with relationships between electricity and chemical reactions. • Consider the spontaneous reaction that occurs when Zn is added to HCl. Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g) • The oxidation numbers of Zn and H+ have changed. • The oxidation number of Zn has increased from 0 to +2. • The oxidation number of H has decreased from +1 to 0. • Therefore, Zn is oxidized to Zn2+, while H+ is reduced to H2. • H+ causes Zn to be oxidized. • Thus, H+ is the oxidizing agent, or oxidant. • Zn causes H+ to be reduced. • Thus, Zn is the reducing agent, or reductant. • Note that the reducing agent is oxidized and the oxidizing agent is reduced. 2. Balancing Oxidation-Reduction Equations • Recall the law of conservation of mass: The amount of each element present at the beginning of the reaction must be present at the end. • Conservation of charge state that electrons are not lost in a chemical reaction. • Some redox equations may be easily balanced by inspection. • However, for many redox reactions we need to look carefully at the transfer of electrons. Half-Reactions • Half-reactions are a convenient way of separating oxidation and reduction reactions. • Consider the reaction: Sn2+(aq) + 2 Fe3+(aq) Sn4+(aq) + 2 Fe2+(aq) • The oxidation half-reaction is: Sn2+(aq) Sn4+(aq) +2 e–1 • Note that electrons are a product here. • The reduction half-reaction is: 2 Fe3+(aq) + 2e– 2 Fe2+(aq) • Note that electrons are a reactant here. Balancing Equations by the Method of Half-Reactions • Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple). • MnO4– is reduced to Mn2+ (pale pink), while the C2O42– is oxidized to CO2. • The equivalence point is indicated by the presence of a pale pink color. • If more KMnO4 is added, the solution turns purple due to the excess KMnO4. • What is the balanced chemical equation for this reaction? • We can determine this using the method of half-reactions: • Write down the two incomplete half reactions. MnO4– (aq) 2+ (aq) C2O42– (aq) CO2(g) • Balance each half reaction. • First, balance elements other than H and O. MnO4–1 (aq) Mn2+(aq) C2O42– (aq) 2 CO2(g) • Then balance O by adding water. MnO4–1 (aq) Mn2+(aq) + 4 H2O(l) C2O42– (aq) 2 CO2(g) • Then balance H by adding H+1. 8 H+1 (aq) + MnO4–1 (aq) Mn2+(aq) + 4 H2O(l) C2O42– (aq) 2 CO2(g) • Finish by balancing charge by adding electrons. • This is an easy place to make an error! • For the permanganate half-reaction, note that there is a charge of 7+ on the left and 2+ on the right. • Therefore, 5 electrons need to be added to the left: 5e–1 + 8 H+1(aq) + MnO4–1 (aq) Mn2+(aq) + 4 H2O(l) • In the oxalate half-reaction, there is a 2– charge on the left and a 0 charge on the right, so we need to add two electrons to the products: C2O42– (aq) 2 CO2(g) + 2 e–1 • Multiply each half-reaction to make the number of electrons equal. • To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. • Multiplying gives: 10e– + 16H+(aq) + 2MnO4– (aq) 2Mn2+(aq) + 8H2O(l) 5C2O42– (aq) 10CO2(g) + 10e– • Now add the reactions and simplify. 16 H+1(aq) + 2 MnO4–1 (aq) + 5 C2O42– (aq) 2 Mn2+(aq) + 8 H2O(l) + 10 CO2(g) • The equation is now balanced! • Note that all of the electrons have cancelled out ! Balancing Equations for Reactions Occurring in Basic Solution • The same method as above is used, but OH– is added to “neutralize” the H+ used. • The equation must again be simplified by canceling like terms on both sides of the equation. 3. Voltaic Cells • The energy released in a spontaneous redox reaction may be used to perform electrical work. • Voltaic, or galvanic cells, are devices in which electron transfer occurs via an external circuit. • Voltaic cells utilize spontaneous reactions. • If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) • Zn is spontaneously oxidized to Zn2+ by Cu2+. • The Cu2+ is spontaneously reduced to Cu0 by Zn. • The entire process is spontaneous. • Each of the two compartments of a voltaic cell is called a half-cell. • This voltaic cells consists of: • An oxidation half-reaction: Zn(s) Zn2+(aq) + 2e– • Oxidation takes place at the anode. • A reduction half-reaction: Cu2+(aq) + 2e– Cu(s) • Reduction takes place at the cathode. • A salt bridge is used to complete the electrical circuit. • Cations move from anode to cathode. • Anions move from cathode to anode. • The two solid metals are the electrodes (cathode and anode). • As oxidation occurs, Zn is converted to Zn2+ and 2e–. • The electrons flow toward the cathode where they are used in the reduction reaction. • We expect the Zn electrode to lose mass and the Cu electrode to gain mass. • Electrons flow from the anode to the cathode. • Therefore, the anode is negative and the cathode is positive. • Electrons cannot flow through the solution; they have to be transported through an external wire. • Anions and cations move through a porous barrier or a salt bridge. • Cations move into the cathodic compartment to neutralize the excess of negatively charged ions (Cathode: Cu2+ + 2e– Cu, so the counter ion of Cu is in excess). • Anions move into the anodic compartment to neutralize the excess Zn2+ ions formed by oxidation. A Molecular View of the Electrode Process • “Rules” of voltaic cells: • At the anode electrons are products. • Oxidation occurs at the anode. • At the cathode electrons are reagents. • Reduction occurs at the cathode. • The flow of electrons from anode to cathode requires an external wire. • The transfer of ions through a salt bridge maintains overall charge balance for the two compartments. 4. Cell EMF Under Standard Conditions • The flow of electrons from anode to cathode is spontaneous. • What is the “driving force”? • Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode. • Potential difference is the difference in electrical potential. • The potential difference is measured in volts. • One volt (V) is the potential difference required to impart one joule (J) of 1V 1 J C energy to a charge of one coulomb (C): • Electromotive force (emf) is the force required to push electrons through the external circuit. • Cell potential: Ecell is the emf of a cell. • This is known as the cell voltage. • Ecell is > 0 for a spontaneous reaction. • For 1M solutions or 1 atm pressure for gases at 25 the standard emf (standard cell potential) is called • For example, for the reaction: conditions), E cell. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) • E cell = +1.10 V Standard Reduction (Half-Cell) Potentials • We can conveniently tabulate electrochemical data. • Standard reduction potentials, E are measured relative to a standard. • The emf of a cell can be calculated from standard reduction potentials: E = (cathode) – E (anode) • We use the following half-reaction as our standard: 2H+ (aq, 1M) + 2e– H2(g, 1 atm) E cell = 0V. • This electrode is called a standard hydrogen electrode (SHE) or the normal hydrogen electrode (NHE). • The SHE is assigned a standard reduction potential of zero. • Consider the half-reaction: Zn(s) Zn2+(aq) + 2 e–1 • We can measure E relative to the SHE (cathode): • It consists of a Pt electrode in a tube placed in 1 M H+ solution. • H2 is bubbled through the tube. E = E (cathode) – E (anode) 0.76 V = 0 V – E (anode). • Therefore, E (anode) = – 0.76 V. • Standard reduction potentials must be written as reduction reactions: Zn2+(aq, 1M) + 2 e–1 Zn(s) E = –0.76 V. • Since E = –0.76 V we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous. • However, the oxidation of Zn with the SHE is spontaneous. • The standard reduction potential is an intensive property. • Therefore, changing the stoichiometric coefficient does not affect E . 2 Zn2+(aq) + 4e– 2 Zn(s) E = –0.76 V • Reactions with E > 0 are spontaneous reductions relative to the SHE. • Reactions with E < 0 are spontaneous oxidations relative to the SHE. • The larger the difference between E values, the larger E . • The more positive the value of E , the greater the driving force for reduction. Strengths of Oxidizing and Reducing Agents • Consider a table of standard reduction potentials. • We can use this table to determine the relative strengths of reducing (and oxidizing) agents. • The more positive the E , the stronger the oxidizing agent (written in the table as a reactant). • The more negative the E , the stronger the reducing agent (written as a product in the table). • We can use this to predict if one reactant can spontaneously oxidize another. • For example: • F2 can oxidize H2 or Li. • Ni2+ can oxidize Al(s). • We can use this table to predict if one reactant can spontaneously reduce another. For example: Li can reduce F2. 5. Free Energy and Redox Reactions • For any electrochemical process E E (reduction process) – E (oxidation process). • A positive E • A negative E • The above equation is used to understand the activity series of metals. • Consider the reaction of nickel with silver ion: Ni(s) + 2Ag+(aq) Ni2+(aq) + 2Ag(s) • The standard cell potential is: E =E (Ag+/Ag) – E (Ni2+/Ni) = (0.80 V) – (–0.28 V) = 1.08 V • This value indicates that the reaction is spontaneous. G • We can show that: G nFE • G is the change in free energy, n is the number of moles of electrons transferred, F is Faraday's constant, and E is the emf of the cell. • We define a faraday (F) as: 1F 96,485 C J 96,485 mol V - mol • Since n and F are pos G < 0 then E > 0 and the reaction will be spontaneous. • When the reactants and products are in their standard states: Go = –nFEo 6. Cell EMF Under Nonstandard Conditions • A voltaic cell is functional until E = 0 at which point equilibrium has been reached. • The cell is then “dead.” • The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. The Nernst Equation • We can calculate the cell potential under nonstandard conditions. • Recall that: G G RT lnQ • We can substitute in our expression for the free energy change: nFE nFE RTlnQ • Rearranging, we get the Nernst equation: or • Note that there is a change from natural logarithm to log base 10. • The Nernst equation can be simplified by collecting all the constants together 2.303 RT log Q nF E E and using a temperature of 298 K: 0.0592 E E log Q n • Example: If we have the reaction: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) • If [Cu2+] = 5.0 M and [Zn2+] = 0.050M: Concentration Cells • A concentration cell is one whose emf is generated solely because of a concentration difference. • Example: Consider a cell with two compartments, each with a Ni(s) electrode but with different concentrations of Ni2+(aq). • One cell has [Ni2+] = 1.0 M and the other has [Ni2+] = 0.001 M. • The standard cell potential is zero. • But this cell is operating under nonstandard conditions! • The driving force is the difference in Ni2+ concentrations. • Anode (dilute Ni2+): Ni(s) Ni2+(aq) + 2e • Cathode (concentrated Ni2+): Ni2+(aq) + 2e Ni(s) • Using the Nernst equation we can calculate a cell potential of +0.0888 V for this concentration cell. 7. Batteries and Fuel Cells • A battery is a portable, self-contained electrochemical power source consisting of one or more voltaic cells. Lead-Acid Battery • A 12 V car battery consists of six cathode/anode pairs each producing 2 V. • Cathode: PbO2 on a metal grid in sulfuric acid: PbO2(s) + HSO4 -1 (aq) + 3 H+1(aq) + 2e 1 PbSO4(s) + 2H2O(l) • Anode: Pb: Pb(s) + HSO4-1 (aq) PbSO4(s) + H +1 (aq) + 2 e-1 • The overall electrochemical reaction is PbO2(s) + Pb(s) + 2 HSO4-1 (aq) + 2 H+1 (aq) 2 PbSO4(s) + 2 H2O(l) • The cell potential for this reaction is: E =E E (anode) = +2.041 V • Wood or glass-fiber spacers are used to prevent the electrodes from touching. • An advantage of these cells is that they can be recharged. • An external source of energy is used to reverse the process. Alkaline Battery • The most common nonrechargeable battery is the alkaline battery. • Powdered zinc metal is immobilized in a gel in contact with a concentrated solution of KOH. • Thus, these batteries are alkaline. • The reaction at the anode is: Zn(s) + 2 OH-1 (aq) Zn(OH)2(aq) + 2e • The reaction at the cathode is the reduction of MnO2: 2 MnO2(s) + 2 H2O(l) + 2e 2 MnO(OH)(s) + 2OH (aq) • The cell potential of these batteries is 1.55 V at room temperature. Nickel-Cadmium, Nickel-Metal-Hydride, and Lithium-Ion Batteries • A common rechargeable battery is the nickel-cadmium (NiCad) battery. • The reaction at the cathode is: 2NiO(OH)(s) + 2H2O(l) + 2e 2Ni(OH)2(s) +2OH (aq) • The reaction at the anode is: Cd(s) + 2OH (aq) Cd(OH)2(s) + 2e • The cell potential of this battery is about 1.30 V at room temperature. • Cadmium is a toxic heavy metal. • There are environmental concerns to be addressed regarding the disposal of such batteries. • Other rechargeable batteries have been developed. • NiMH batteries (nickel-metal-hydride). • Li-ion batteries (lithium-ion batteries). Hydrogen Fuel Cells • Direct production of electricity from fuels occurs in a fuel cell. • An example is a hydrogen fuel cell. • At the cathode: 2 H2O(l) + O2(g) + 4 e-1 4 OH -1(aq) • At the anode: 2 H2(g) + 4 OH-1 (aq) 4 H2O(l) + 4e • This cell is known as a PEM fuel cell (proton exchange membrane). • The anode and cathode are separated by a polymer membrane that is permeable to protons but not electrons. • It acts as a salt bridge. Direct Methanol Fuel Cells • These fuel cells are similar to the PEM cell but use methanol as a reactant instead of hydrogen gas. 8. Corrosion • An example of an undesirable redox reaction is the corrosion of metals. • Metal is attacked by a substance in the environment and converted to an unwanted compound. Corrosion of Iron • Consider the rusting of iron: • Since E (Fe2+) < E (O2), iron can be oxidized by oxygen. • Cathode: O2(g) + 4 H+(aq) + 4 e 2H2O(l) E = 1.23 V. • Anode: Fe(s) Fe2+(aq) + 2e E • Dissolved oxygen in water usually causes the oxidation of iron. • The Fe2+ initially formed can be further oxidized to Fe3+, which forms rust, Fe2O3.x H2O(s). • Oxidation occurs at the site with the greatest concentration of O2. • Other factors to consider are the pH, presence of salts, stress on the iron, and contact with other metals. Preventing the Corrosion of Iron • Corrosion can be prevented by coating the iron with paint or another metal. • This prevents oxygen and water from reacting at the surface of the iron. • Galvanized iron is coated with a thin layer of zinc. • Zinc protects the iron since Zn is the anode and Fe is the cathode: Zn2+(aq) +2e Zn(s) E 2+ Fe (aq) + 2e Fe(s) E • The standard reduction potentials indicate that Zn is easier to oxidize than Fe. • This process is cathodic protection (the sacrificial anode is destroyed). • We can use something similar to protect underground pipelines. • Often, Mg is used as a sacrificial anode: Mg2+(aq) +2 e - Mg(s) E Fe2+(aq) + 2 e - Fe(s) E 9. Electrolysis • Electrolysis reactions are nonspontaneous reactions that require an external current in order to force the reaction to proceed. • They take place in electrolytic cells. • In voltaic and electrolytic cells, reduction occurs at the cathode and oxidation occurs at the anode. • However, in electrolytic cells, electrons are forced to flow from the anode to the cathode. • In electrolytic cells the anode is positive and the cathode is negative. • In voltaic cells the anode is negative and the cathode is positive. • Example: The decomposition of molten NaCl. • Cathode: 2 Na+(l) + 2e 2Na(l) • Anode: 2 Cl (l) Cl2(g) + 2e -1. • Industrially, electrolysis is used to produce metals like Al. • Electrolysis of high-melting ionic substances requires very high temperatures. • Do we get the same products if we electrolyze an aqueous solution of the salt? • Water complicates the issue! • Example: Consider the electrolysis of NaF(aq): Na+ (aq) + e-1 Na(s) E 2 H2O(l) + 2 e -1 H2(g) + 2 OH-1 (aq) E • Thus, water is more easily reduced than the sodium ion. 2 F (aq) F2(g) + 2 e -1 E = +2.87 V 2 H2O(l) O2(g) + 4 H+1(aq) + 4 e -1 E = +1.23 V • Thus, it is easier to oxidize water than the fluoride ion. • Electrolysis does not always involve inert electrodes. • Active electrodes are electrodes that take part in electrolysis. • An example is electroplating. • Consider an active Ni electrode and another metallic electrode (steel) placed in an aqueous solution of NiSO4: • Anode (nickel strip): Ni(s) Ni2+(aq) + 2 e -1 • Cathode (steel strip): Ni2+(aq) + 2 e-1 Ni(s) • Ni plates on the inert electrode. • Electroplating is important in protecting objects from corrosion. Quantitative Aspects of Electrolysis • We want to know how much material we obtain with electrolysis. • Consider the reduction of Cu2+ to Cu. Cu2+(aq) + 2e Cu(s). • 2 mol of electrons will plate 1 mol of Cu. • The charge of one mol of electrons is 96,500 C (1 F). • A coulomb is the amount of charge passing a point in one second when the current is one ampere. • The amount of Cu can be calculated from the current (I) and time (t) required to plate. Q=It Electrical Work • Free energy is a measure of the maximum amount of useful work that can be obtained from a system. • We know: G = wmax and: G nFE • thus: wmax nFE • If Ecell is positive, wmax will be negative. • Work is done by the system on the surroundings. • The emf can be thought of as being a measure of the driving force for a redox process. • In an electrolytic cell an external source of energy is required to force the reaction to proceed. w = nFEexternal • In order to drive the nonspontaneous reaction, the external emf must be greater than Ecell. • From physics we know that work is measured in units of watts: 1 W = 1 J/s • Electric utilities use units of kilowatt-hours: 3600s 1 J/s 6 1 kWh 1000 W1 hour 3.6 10 J 1 hour 1 W